Mechanics of Machine

MEEM 3700 – Suggested problems from Chapter 1 1.7.1 Given that ω=5 rad/sec, then the frequency in Hz is f=ω/2π and the period T= 1/f 1.15 The displacement amplitude X=0.15mm or 0.00015m The acceleration amplitude is ω2X = 0.6g = 0.6 (9.81) m/sec2 ω2 = ω2X/ X ; but ω is in rad/sec. Convert rad/sec to rpm to get the shaft speed (1890 rpm). 1.18 p=polyfit(x,f,1) ; p=35.18, -0.0607 and f=35.18x - 0.0607; k=35.18 (the slope of the line) 1.19 For springs end to end (series), 1/keq = 1/k1 = 1/k2 ; for equal springs, the equivalent spring constant is 1/keq = 1/k + 1/k = 2/k and ker = k/2 (weaker) For springs side-by-side (parallel ) keq = k1 + k2 and keq = 2k (twice as stiff) 1.20 1.21 Assume “thickness” is the beam height h. K = f/x, for the doubly clamped beam (both ends clamped), k = f/x = 192EI/L3 and I = 1/12 bh3 Take moments around the pivot point O. The force in each spring is the product of its spring constant (k) and its deflection. fL = (kx)L + (K x/2)L/2 = KLx(5/4) The equivalent spring constant is Keq= f/x = 5K/4

1.28

1.38

Take moments about the pivot, the restoring torque is T = (mg)(Lsinθ) For small angles, sinθ is approximately equal to θ in radians and KT = T/θ = mgL

1.39

The buoyant force is equal to the weight of the water displaced = vol x density. F = (πD2/4)(x)(ρg) ; k=F/x=(πD2/4)ρg

Use MATLAB to check the spring constant obtained from the slope of the f vs. x plot. Problem 1.49

1.49

x=[0,.3,.46,.7,.75,1.05,1.14,1.36,1.55]; f=[0:200:1600] p=polyfit(x,f,1) xfit=[0:.1:1.5] ffit=polyval(p,xfit) plot(x,f,'o',xfit,ffit),xlabel('x-cm'),ylabel('f-N')

1600 1400 1200 1000 800 600 400 200 0 -200

f-N

0

0.2

0.4

0.6

0.8 x - cm

1

1.2

1.4

1.6

1.59

x=[0.03:0.03:0.18] f=[45,102,165,245,360,512] xs=[-x,0,x] fs=[-f,0,f] p=polyfit(xs,fs,3)

(if “0” is included, it will wind up being put in 2 times) (makes symmetric about origin)